Question:

**Step 1:** Convert the logarithms to a common base, base 3, using the change of base formula $\log_x y = \frac{\log_z y}{\log_z x}$. We have: $$ \log_{27} a = \frac{\log_3 a}{\log_3 27} = \frac{\log_3 a}{3} $$ $$ \log_9 b = \frac{\log_3 b}{\log_3 9} = \frac{\log_3 b}{2} $$ $$ \log_{27} b = \frac{\log_3 b}{\log_3 27} = \frac{\log_3 b}{3} $$ $$ \log_9 a = \frac{\log_3 a}{\log_3 9} = \frac{\log_3 a}{2} $$ **Step 2:** Rewrite the given equations using the expressions from Step 1. The first equation is $\log_{27} a + \log_9 b = \frac{7}{2}$: $$ \frac{\log_3 a}{3} + \frac{\log_3 b}{2} = \frac{7}{2} $$ Multiply the entire equation by 6 to clear the denominators: $$ 2\log_3 a + 3\log_3 b = 21 \quad (*)$$ The second equation is $\log_{27} b + \log_9 a = \frac{2}{3}$: $$ \frac{\log_3 b}{3} + \frac{\log_3 a}{2} = \frac{2}{3} $$ Multiply the entire equation by 6 to clear the denominators: $$ 2\log_3 b + 3\log_3 a = 4 \quad (**)$$ **Step 3:** Solve the system of linear equations for $\log_3 a$ and $\log_3 b$. The system is: 1. $2\log_3 a + 3\log_3 b = 21$ 2. $3\log_3 a + 2\log_3 b = 4$ Multiply equation (1) by 3 and equation (2) by 2 to eliminate $\log_3 a$: $$ 3 \times (2\log_3 a + 3\log_3 b) = 3 \times 21 \implies 6\log_3 a + 9\log_3 b = 63 $$ $$ 2 \times (3\log_3 a + 2\log_3 b) = 2 \times 4 \implies 6\log_3 a + 4\log_3 b = 8 $$ Subtract the second new equation from the first new equation: $$ (6\log_3 a + 9\log_3 b) - (6\log_3 a + 4\log_3 b) = 63 - 8 $$ $$ 5\log_3 b = 55 $$ $$ \log_3 b = 11 $$ From $\log_3 b = 11$, we find $b = 3^{11}$. Substitute $\log_3 b = 11$ into equation (1): $$ 2\log_3 a + 3(11) = 21 $$ $$ 2\log_3 a + 33 = 21 $$ $$ 2\log_3 a = 21 - 33 $$ $$ 2\log_3 a = -12 $$ $$ \log_3 a = -6 $$ From $\log_3 a = -6$, we find $a = 3^{-6}$. **Step 4:** Calculate the product $ab$. $$ ab = (3^{-6})(3^{11}) $$ Using the property $x^m \cdot x^n = x^{m+n}$: $$ ab = 3^{-6+11} $$ $$ ab = 3^5 $$ $$ ab = 243 $$